MATHS M.C.Q

Here, given $AB = BC = CD$
Now, equal chords subtend equal angles at centre. So, $\angle\text{AOC}=\angle\text{BOC}=\angle\text{COD}$
Also, they lie in straight line so, $\angle\text{AOC}+\angle\text{BOC}+\angle\text{COD}=180^\circ$
$\angle\text{AOB}=\angle\text{BOC}=\angle\text{COD}=60^\circ$
In $\triangle\text{AOB}$
$\text{AO}=\text{OB},\angle\text{OAB}=\angle\text{OBA}$
Since, $\angle\text{AOB}=60^\circ,\angle\text{OAB}=\angle\text{OBA}=60^\circ$
Now, $\angle\text{DOE}=\angle\text{AOB}=60^\circ$ (vertically opposite angle)
In $\triangle\text{DOE},\text{OD}=\text{OE}$ (radius)
so, $\angle\text{ODE}=\angle\text{OED}$
$\triangle\text{DOE},\angle\text{DOE}+\angle\text{ODE}+\angle\text{OED}=180^\circ$
$2\angle\text{ODE}=\angle\text{OED}=180-60=120^\circ$
$\angle\text{ODE}=\angle\text{OED}=60^\circ$
given was, $\angle\text{DEF}=110^\circ,$ so, $\angle\text{OEF}=110-60=50^\circ$
Now, in $\triangle\text{EOF}\ \text{OE}=\text{OF}$
so, $\angle\text{OEF}=\angle\text{OFE}=50^\circ$
In, $\triangle\text{EOF}\ \angle\text{FOE}=180-(50+50)=80^\circ$
Now, $\angle\text{DOE}+\angle\text{FOE}+\angle\text{AOF}=180^\circ$ (All lie on same straight line)
So, $\angle\text{AOF}=180-(80+60)=40^\circ$
Now, in $\triangle\text{AOF}\ \text{AO}=\text{FO}$
SO, $\angle\text{OFA}=\angle\text{OAF}$
In $\triangle\text{AOF},2\angle\text{OAF}+\angle\text{FOA}=180^\circ$
$2\angle\text{OAF}+\angle\text{FOA}=180^\circ$
$\angle\text{OAF}=90-20=70^\circ$
So, $\angle\text{FAB}=\angle\text{FAO}+\angle\text{OAB}$
$=70^\circ+60^\circ=130^\circ$

We know that the opposite angles of a quadrilateral are supplementary.
$\angle\text{ADC}+\angle\text{ABC}=180^\circ$
$\Rightarrow\ 120^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\ \angle\text{ABC}=60^\circ$
Since BOC is a diameter $\angle\text{ACB}=90^\circ.$
In $\triangle\text{CAB},$
$\angle\text{ABC}+\angle\text{BAC}+\angle\text{ACB}=180^\circ$ [Angle sum property]
$\Rightarrow\ 60^\circ+\angle\text{BAC}+90^\circ=180^\circ$
$\Rightarrow\ \angle\text{BAC}=30^\circ$

In $\triangle\text{OAB},$ we have:
$OA = OB ($Radii of a circle$)$
$\Rightarrow\angle\text{OAB}=\angle\text{OBA}=20^\circ$
In $\triangle\text{OAC},$ we have:
$OA = OC ($Radii of a circle$)$
$\Rightarrow\angle\text{OAC}=\angle\text{OCA}=30^\circ$
Now, $\angle\text{BAC}=(20^\circ+30^\circ)=50^\circ$
$\therefore\angle\text{BOC}=(2\times\angle\text{BAC})=(2\times50^\circ)=100^\circ$
$\Rightarrow\angle\text{BOC}=100^\circ$

Equal arcs subtend equal angles at the centre and the angle subtended by them at the circumference would be double the angle subtended by them at the centre. As the angle subtended at centre were same so the angle subtended at the circumference would also become same. Thus each arc would make an angle of $19^\circ .$ Thus the total length of all the three angles would be thrice $19^\circ$ that is $57^\circ .$

Angle made by a chord at the centre is twice the angle made by it on any point on the circumference. Therefore,
$\angle\text{QOR}=2\angle\text{QPR}=50^\circ\times2=100^\circ$

$\angle\text{OQP}=\text{OPQ}=50^\circ$
$\angle\text{POQ}=80^\circ$ (From angle sum property)
$\angle\text{SOQ}=180^\circ-80^\circ=100^\circ$ (From linear pair)
Completing the cyclic quadrilateral, $QRSL, (L$ being any point on the circumference$)$
$\angle\text{SLQ}=50^\circ$
From cyclic quadrilateral, we have
$\angle\text{SRQ}=180^\circ-50^\circ=130^\circ$

$OA = OB ($Radii of a circle$)$
$\Rightarrow\angle\text{OBA}=\angle\text{OAB}=50^\circ$
In $\triangle\text{OAB},$ we have:
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ( $Angle sum property of a triangle$)$
$\Rightarrow50^\circ+50^\circ+\angle\text{AOB}=180^\circ$
$\Rightarrow\angle\text{AOB}=(180^\circ-100^\circ)=80^\circ$
Since $\angle\text{AOB}+\angle\text{BOD}=180^\circ$ (Linear pair)
$\therefore\angle\text{BOD}=(180^\circ-80^\circ)=100^\circ$
$\Rightarrow\angle\text{BOD}=100^\circ$

Since $ABCD$ is a cyclic quadrilateral
$\angle\text{B}+\angle\text{D}=180^\circ$
$60^\circ+\angle\text{D}=180^\circ$
$\angle\text{D}=120^\circ$
Now since $AD$ is parallel to $BC$
$\angle\text{C}+\angle\text{D}=180^\circ$
$\angle\text{C}+120^\circ=180^\circ$
$\angle\text{C}=60^\circ$

Here we have a cyclic quadrilateral $PQRS$ with $PR$ being a diameter of the circle. Let the centre of this circle be $O.$
We are given that $\angle\text{QPR}$ and $\angle\text{SPR}=72^\circ.$

So, we see that,
$\angle\text{QPS}=\angle\text{QPR}+\angle\text{RPS}$
$=67^\circ+72^\circ$
$=139^\circ$
In a cyclic quadrilateral, it is known that the opposite angles as supplementary.
$\angle\text{QPS}+\angle\text{QRS}=180^\circ$
$\angle\text{QRS}=180^\circ-\angle\text{QPS}$
$\angle\text{QRS}=180^\circ-139^\circ=41^\circ$
$\angle\text{QRS}=41^\circ$

In $\triangle\text{QAB, OA} = \text{OB}$ [both are the radius of a circle]
$\angle\text{OAB} = \angle\text{OBA}\Rightarrow \angle\text{OBA} = 40^\circ$
[angles opposite to equal sides are equal]
Also, $\angle\text{AOB} = \angle\text{OBA}\Rightarrow \angle\text{BAO} = 180^\circ$
[by angle sum property of a triangle]
$\angle\text{AOB} + 40^\circ + 40^\circ = 180^\circ$
$\Rightarrow\ \angle\text{AOB} = 180^\circ – 80^\circ = 100^\circ$
We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
$\angle\text{AOB} = 2 \angle\text{ACB} \Rightarrow 100^\circ =2 \angle\text{ACB}$
$\angle\text{ACB} = \frac{100}{2} = 50^\circ$

We are given that the chord is equal to its radius.
We have to find the angle subtended by this chord at the minor arc.
We have the corresponding figure as follows:

We are given that
$AO = OB = AB$
So,
$\triangle\text{AOB}$ is an equilateral triangle.
Therefore, we have
$\angle\text{AOB}=60^\circ$
Since the angle subtended by any chord at the centre is twice the angle subtended at any point on the circle.
So, $\angle\text{AQB}=\frac{\angle\text{AOB}}{2}=\frac{60}{2}=30^\circ$
$\Rightarrow\angle\text{AQB}=30^\circ$
Take a point $P$ on the minor arc
Since $APBQ$ is a cyclic quadrilateral
So, opposite angles are supplementary. That is
$\angle\text{APB}+\angle\text{AQB}=180^\circ$
$\angle\text{APB}+130^\circ=180^\circ\ [\angle\text{AQB}=30^\circ]$
$\angle\text{APB}=180^\circ-30^\circ=150^\circ$

Since the chord is equal to the radius therefore, it will form an equilateral triangle inside the circlewith the third vertex being the centre of the circle.
So the chord will make an angle of $60^\circ $ at the centre. As the angle made by the chord at any other point of the circumfrence would be half.
So, we have that angle made at the major segment would be $30^\circ .$

$OA = OB [$Radii of the same circle$]$
$\Rightarrow\ \angle\text{OAB}=\angle\text{OBA}=50^\circ [$Angle opposite equal sides are equal$]$
$\angle\text{BOD}=\angle\text{OAB}+\angle\text{OBA}$
$\Rightarrow\ \angle\text{BOD}=50^\circ+50^\circ$
$\Rightarrow\ \angle\text{BOD}=100^\circ$

$ABC$ is an arc of circle.
Take point $D$ in the altrenative segment and join $AD$ and $CD.$
$\angle\text{ABC}=135^\circ$ (Given)
$\angle\text{ABC}+\angle\text{ADC}=180^\circ ($Sum of opposite angles of cyclic quadrilateral is $180^\circ )$
$\Rightarrow\angle\text{ADC}=180^\circ-\angle\text{ABC}=180^\circ-135^\circ=45^\circ$
Now, $\angle\text{AOC}=2\times\angle\text{ADC}=2\times45^\circ=90^\circ$
$\widehat{\text{ABC}}=$ measure of the central angle $=\angle\text{AOC}=90^\circ$
$\Rightarrow\text{Required ratio}=\frac{\text{arc}\widehat{\text{ABC}}}{\text{circumference}}$
$=\frac{90^\circ}{360^\circ}=\frac{1}{4}=1:4$

$\angle\text{BAC}=60^\circ$ (Angle of equilateral triangle)
Arc $\widehat{\text{BC}}$ makes angle $\angle\text{BAC}$ at circle and $\angle\text{BOC}$ at center of circle.
$\Rightarrow\angle\text{BAC}=\frac{1}{2}\angle\text{BOC}$
$\Rightarrow2\times\angle\text{BAC}=\angle\text{BOC}$
$\Rightarrow2\times60^\circ=\angle\text{BOC}$
$\Rightarrow\angle\text{BOC}=120^\circ$

In, $\triangle\text{ABD}$
$\angle\text{D}=180^\circ-\angle\text{A}-\angle\text{B}$
$=180^\circ-110^\circ=70^\circ$
Since angles made by same chord at any point of circumference are equal so, $\angle\text{ACB}=\angle\text{ADB}=70^\circ$

Since, $PX = XQ$
$2PX = PQ$
$PX : PQ = 1 : 2$

The angle in a semicircle measures $90^\circ .$

Since $ABCD$ is a cyclic qyadrilateral, we have:
$\angle\text{BAD}+\angle\text{BCD}=180^\circ$
$\Rightarrow\ \angle\text{BAD}+110^\circ=180^\circ$
$\Rightarrow\ \angle\text{BAD}=70^\circ$
Since $ABEF$ is a cyclic qyadrilateral, we have:
$\angle\text{BAD}+\angle\text{BEF}=180^\circ$
$\Rightarrow\ 70^\circ+\angle\text{BEF}=180^\circ$
$\Rightarrow\ \angle\text{BEF}=110^\circ$

Since the chord is equal to the radius therefore, it will form an equilateral triangle inside the circle with the third vertex being the centre of the circle.
So the chord will make an angle of $60^\circ $ at the centre. As the angle made by the chord at any other point of the circumference would be half.
So, we have that angle made at the major segment would be $30^\circ .$

In $\triangle\text{OAB},$
$OA = OB [$Radii of the same circle$]$
$\Rightarrow\ \angle\text{OBA}=\angle\text{OAB}=20^\circ[ $Angle opposite equal sides are equal$]$
In $\triangle\text{OAC},$
$OA = OC [$Radii of the same circle$]$
$\Rightarrow\ \angle\text{OCA}=\angle\text{OAC}=30^\circ [$Angle opposite equal sides are equal$]$
Now, $\angle\text{BAC}=\angle\text{BAO}+\angle\text{CAO}$
$=20^\circ+30^\circ$
$=50^\circ$
$\angle\text{BOC}=2\angle\text{BAC}=2(50^\circ)=100^\circ.$

Angle made by a chord at the centre is twice the angle made by it on any point on the circumference. Therefore,
$\angle\text{QOR}=2\angle\text{QPR}=50^\circ\times2=100^\circ$

In triangle $ABO, AO = BO$
So, $\angle\text{BAO}=\angle\text{ABO}=\text{x}$
$\text{x}+\text{x}+140^\circ=180^\circ$
$\Rightarrow2\text{x}=40^\circ$
$\text{x}=20^\circ$
Now $\angle\text{QAO}+\angle\text{BAO}+\angle\text{PAB}=180^\circ$
Substituting the values we get:
$90^\circ+20^\circ+\angle\text{PAB}=180^\circ$
$\angle\text{PAB}=70^\circ$

$BC = BD [$Given$]$
$\angle\text{BDC}=\angle\text{CBD}=35^\circ$ [Angle opposite equal sides are equal]
In $\triangle\text{BCD},$
$\angle\text{BCD}+\angle\text{BDC}+\angle\text{CBD}=180^\circ$ [Angle sum property]
$\Rightarrow\ \angle\text{BCD}+35^\circ+35^\circ=180^\circ$
$\Rightarrow\ \angle\text{BCD}=110^\circ$
Since $ABCD$ is a cyclic quadrilateral,
$\angle\text{BAD}+\angle\text{BCD}=180^\circ$
$\Rightarrow\ \angle\text{BAD}+110^\circ=180^\circ$
$\Rightarrow\ \angle\text{BAD}=70^\circ$

We have:
$\angle\text{ABC}+\angle\text{ADC}=180^\circ$
$\Rightarrow\angle\text{ABC}+95^\circ=180^\circ$
$\Rightarrow\angle\text{ABC}=(180^\circ-95^\circ)=85^\circ$
Now, $CF || AB$ and $CB$ is the transversal.
$\therefore\angle\text{BCF}=\angle\text{ABC}=85^\circ($Alternate interior angles$)$
$\Rightarrow\angle\text{BCE}=(85^\circ+20^\circ)=105^\circ$
$\Rightarrow\angle\text{DCB}=(180^\circ-105^\circ)=75^\circ$
$\Rightarrow\angle\text{DCB}=75^\circ$
Now, $\angle\text{BAD}+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{BAD}+75^\circ=180^\circ$
$\Rightarrow\angle\text{BAD}=(180^\circ-75^\circ)$
$\Rightarrow\angle\text{BAD}=105^\circ$

Given that $AB = CD.$
Since equal chord, subtend equal angles at the centre,
$\angle\text{COD}=\angle\text{AOB}=80^\circ.$